Matter & Motion 2000

Chemistry Homework Answer Key #3

Chapter 9 - Spontaneous Processes and Thermodynamic Equilibrium
 

17.

CS2 (g) + 3 O2 (g) ® CO2 (g) + 2 SO2 (g)                K = K1

1/3 CS2 (g) + O2 (g) ® 1/3 CO2 (g) + 2/3 SO2 (g)                K = (K1)1/3

This is an example of a reaction for which the stoichiometric coefficents have all been multiplied by a constant, n. In all such cases the equilibrium for the "new" reaction is related to the stoichiometry of the "old" reaction by, Knew = (Kold)n.
 

19. The reaction for which we are trying to find an equilibrium constant is equal to the sum of the second reaction plus the reverse of the first. Thus, the overall equilibrium constant would be is equal to the product of the individual equilibrium constants, K2/K1.
 
 
XeO4 (g) + XeF6 (g) ® XeOF4(g) + XeO3F2 (g)
K = K2
2 HF (g) + XeOF4 (g) ® XeF6(g) + H2O (g)
K = 1/K1
 ________________________________________
 _____________
XeO4 (g) + 2 HF (g) ® H2O (g) + XeO3F2 (g)
K = K2/K1

 

23. For the reaction, PCl5 (g) « PCl3 (g) + Cl2 (g), the equilibrium expression is:

To calculate the equilibrium concentrations, a table of partial pressures is set-up.
 
 
PCl5
PCl3
Cl2
initial pressure (atm)
0.895
0
0
D
-x
x
x
equilibrium pressure (atm)
0.895 - x
x
x

Substituting into the equilibrium expression yields,

Rearranging yields the quadratic, 0 = x2 + 2.15x + 0.192. Solving for x yields two values:

x = 0.6799, -2.829

Only the first value makes physical sense (the value of xin this problem can not be negative). The partial pressures of the gases are then,

PPCl5 = 0.215 atm

PPCl3 = 0.680 atm

PCl2 = 0.680 atm


25. For the reaction, Br2 (g) + I2 (g) « 2 BrI (g), the equilibrium expression is:

To calculate the equilibrium concentrations, a table of concentration values is set-up.
 
 
Br2
I2
IBr
initial pressure (atm)
0.0500
0.0400
0
D
-x
-x
+2x
equilibrium pressure (atm)
0.0500 - x
0.0400 - x
2x

Substituting into the equilibrium expression yields,

Rearranging yields the quadratic, 0 = 318 x2 -28.98 x + 0.644. Solving for x yields two values:

x = 0.03842, 0.05271

Only the first value makes physical sense (the value of x can not exceed the initial partial pressure of I2, which is 0.04 atm). The partial pressures of the gases at equilibrium are then,

PBr2 = 0.0116 atm

PI2 = 0.00160 atm

PIBr = 0.0768 atm


27. For the reaction, N2 (g) + O2 (g) « 2 NO (g), the equilibrium expression is:

To calculate the equilibrium concentrations, a table of concentration values is set-up.
 
 
N2
O2
NO
initial pressure (atm)
0.41
0.59
0.22
D
x
x
-2x
equilibrium pressure (atm)
0.41+ x
0.59+ x
0.22 - 2x

Yuck. Given the extremely small value of K, it is apparent that almost all of the NO will react to form N2 and O2. So, to make the math more tractable (as it stands x will be very, very close to 0.11, so close that your calculator would not be able to display the value to enough significant figures for you to see that it is not 0.11). To make life a tiny bit easier, let's assume that all of the NO does react, then x = 0.11 and the partial pressures of N2 and O2 are 0.52 and 0.70. Now, let's perform the equilibrium calculation using the following table.
 
 
 
N2
O2
NO
initial pressure (atm)
0.52
0.70
0
D
-x
-x
2x
equilibrium pressure (atm)
0.52- x
0.70 - x
2x

Substituting into the equilibrium expression yields,

Rearranging yields the quadratic, 0 = (4 - 4.2 ´ 10-31) x2 + (5.124 ´ 10-31) x - 1.5288 ´ 10-31. Solving for x yields only one positive value:

x = 1.95 ´ 10-16

The equilibrium partial pressures are then:

PNO = 3.90 ´ 10-16 atm

PN2 = 0.52 atm

PO2 = 0.70 atm


33. For the reaction, 3 Al2Cl6 (g) « 2 Al3Cl9 (g), the reaction quotient is:

The equilibrium constant as solved in problem 11 is 1.04 ´ 10-4. Therefore, given these partial pressures, Q > K; as the reaction approaches equilibrium it will move toward the left, that is, there will be a net consumption of Al3Cl9.

37. For the reaction, P4 (g) « 2 P2 (g), the reaction quotient is:



Since Q > K, the reaction will shift to right and generate more reactant. To find the equilibrium partial pressures, we set up the following table:
 
 
 
P4
P2
initial pressure (atm)
5.00
2.00
D
x
-2x
equilibrium pressure (atm)
5.00 + x
2.00 -2x

Rearranging yields the quadratic, 0 = 4 x2 - 8.612 x + 0.94. Solving for x yields two values:

x = 0.115, 2.038

Since x must be in the range, 0 < x < 1.00, we can exclude the second solution. The equilibrium partial pressures are then,

PP2 = 1.77 atm

PP4 = 5.12 atm.

If, after equilibrium is reached, the volume of the container is increased, the equilibrium will shift to the right, resulting in the generation of additional P2. To illustrate, let's assume the volume is doubled; the resulting pressures (by P1V1 = P2V2) before equilibrium is reestablished will then be: PP2 = 0.885 atm and PP4 = 2.56 atm. Applying the definition of Q gives,

Since Q < K, more products must be generated to reach equilibrium.
 

39. a) If additional product is added to the mixture, the equilibrium will shift to the left to partially consume the added product; more reactant will be generated.

b) If the volume is reduced, the reaction will shift to the right to partially offset the additional pressure (the right side has fewer gas molecules and a shift to the right will reduce the total pressure).

c) If the reaction is cooled, the reaction will shift to the right to generate more heat to partially counteract the loss of "heat product". (Note that with exothermic reactions, you can view heat as one of the products; with endothermic reactions, you can view heat as one of the reactants).

d) If an inert substance is added to a mixture, there is no effect if the partial pressures of the other species present remain unchanged. In this case however, the total pressure is remaining constant, therefore the partial pressures of the other species are decreasing. This has the same effect as decreasing the partial pressures by increasing volume - the reaction will shift to the left to offset the decreased total partial pressure of the reacting species.

e) As stated above, the addition of an inert substance will have no effect if the partial pressures of the reacting species are unchanged. That is the case here.
 

47 & 51. Postponed until next week….
 

60. Abbreviating the acid monomer as HA and the dimer as (HA)2, or HAHA, the reaction of interest is

2 HA « (HA)2

If the total pressure of the system at equilibrium is 0.725, we can assign PHA as x and P(HA)2 as 0.725 -x. Substituting into the equilibrium expression yields:


Only one value makes sense (x can not be negative in this case). Therefore:

 
P(HA) = 0.327 atm

P(HA)2 = 0.398 atm

Thus the percentage of acetic acid that is dimerized is: